Question

Please help.

I understood the explanation of the last question I asked... This problem seems a little different.

Answer 1

do the same thing as last time
its almost the same...

Answer 2

I got like..

21e^2x-4e^x+1=0

Then when I solved that I got the wrong answer.

Answer 3

put e^-1 = 1/e ... multiply and get quadratic equaiton on e^x
rest is same ...

Answer 4

\[e ^{2x}-4e^x-21=0\]

Answer 5

let e^x =y
then e^2x = y^2
then you will have quadratic equation.

Answer 6

\[(e^x-7)(e^x+3)=0\]

Answer 7

Can you get it from there?

Answer 8

e^x = 7 & e^x = -3?

Answer 9

Soo x = ln 7?

Answer 10

yes

Answer 11

Thanks for the explanation! I appreciate it :D

Answer 12

yw

Answer 13

you are welcome.